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baldymcgee6 Group Title

Partial Derivatives... posted below

  • one year ago
  • one year ago

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  1. baldymcgee6 Group Title
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    |dw:1351748975295:dw|

    • one year ago
  2. hartnn Group Title
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    partial derivative means u treat other variable as constant. u know this ?

    • one year ago
  3. baldymcgee6 Group Title
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    Right, so we treat y as a constant in the first one..

    • one year ago
  4. Dido525 Group Title
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    Which variable are you taking constant>?

    • one year ago
  5. hartnn Group Title
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    yes.

    • one year ago
  6. baldymcgee6 Group Title
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    |dw:1351749134665:dw|

    • one year ago
  7. hartnn Group Title
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    absolutely! that was my next statement, that u need log diff. here.

    • one year ago
  8. baldymcgee6 Group Title
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    In my experiences with log diff I always had a function in the form of y = f(x)^g(x).... So what do we do here?

    • one year ago
  9. baldymcgee6 Group Title
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    do we write y = ......?

    • one year ago
  10. hartnn Group Title
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    taking log on both sides, u get log y = g(x) log f(x) then use product rule

    • one year ago
  11. hartnn Group Title
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    u'll also need a chain rule for this....

    • one year ago
  12. baldymcgee6 Group Title
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    |dw:1351749340629:dw|

    • one year ago
  13. hartnn Group Title
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    oh dear! the y on left and on right are different, u should have used g or some other letter

    • one year ago
  14. baldymcgee6 Group Title
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    oohhhhhh.. that's where I went wrong... this is off my last quiz :/

    • one year ago
  15. baldymcgee6 Group Title
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    so it'll be g'/g on the left?

    • one year ago
  16. hartnn Group Title
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    yes, after doing product rule , u resubstitute for g

    • one year ago
  17. baldymcgee6 Group Title
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    |dw:1351749800366:dw|

    • one year ago
  18. hartnn Group Title
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    thats right

    • one year ago
  19. baldymcgee6 Group Title
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    |dw:1351749846652:dw|

    • one year ago
  20. hartnn Group Title
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    absolutely correct

    • one year ago
  21. hartnn Group Title
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    same for 2nd question, just replace x with y and y with x

    • one year ago
  22. baldymcgee6 Group Title
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    right... I just messed up on calling the function y... messed with my head.

    • one year ago
  23. baldymcgee6 Group Title
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    thanks again for the insight!

    • one year ago
  24. baldymcgee6 Group Title
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    I'm not going to do the second derivative, I think I can get it... Thanks so much!

    • one year ago
  25. baldymcgee6 Group Title
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    I mean, i'm not going to do it on here.. :)

    • one year ago
  26. hartnn Group Title
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    ok, welcome ^_^

    • one year ago
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